Answer:
Oxidation occurs at the anode: Fe(s) | Fe2+(aq) half cell. (ii) Reduction occurs at the cathode: Ag(s) | Ag+(aq) half cell. Oxidation occurs at the anode: Pt | Sn2+(aq), Sn4+(aq) half cell. (iii) Electrons flow from the anode to the cathode: from the Pt(s) → Ag(s) electrode.
The two half-reactions are...
Ag→Ag+
and...
NO3→NO
Let's start by balancing the first half-reaction...
Ag→Ag+
The amounts are already balanced; 1:1. The oxygens are balanced. So all that's left is to balance the charge...
Ag→Ag++e−
Now let's do the other equation... Amounts of nitrogen are balanced, so we first need to balance the oxygens...
NO3→NO
4H++NO3→NO+2H2O
Next, we need to balance charge...
4e−+4H++NO3→NO+2H2O
Now let's go ahead and rewrite each half-reaction after being balanced by themselves...
Ag→Ag++e−
4e−+4H++NO3→NO+2H2O
Now we need to multiply by some factor to get the electrons to cancel out. In this case, that factor is 4, which needs to be applied to the top half-reaction...
4(Ag→Ag++e−)=4Ag→4Ag++4e−
Then we combine this half-reaction with the second one above to get...
4Ag+4H++NO3→4Ag++NO+2H2O
I don’t know “which substance” are you talking about??
You can put the picture there so we can see.
